Answer:
x = 0.324 M s⁻¹
Explanation:
Equation for the reaction can be represented as:
2 NO(g) + Cl₂ (g) ⇄ 2NOCl (g)
Rate = K [NO]² [Cl₂]
Concentration = ![\frac{numbers of mole (n)}{volume (v)}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumbers%20of%20mole%20%28n%29%7D%7Bvolume%20%28v%29%7D)
from the question; their number of moles are constant since the species are quite alike.
As such; if Concentration varies inversely proportional to the volume;
we have: Concentration ∝ ![\frac{1}{v}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%7D)
Concentration = ![\frac{1}{v}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%7D)
Similarly; the Rate can now be expressed as:
Rate = K [NO]² [Cl₂]
Rate =
![(\frac{1}{v} )](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7Bv%7D%20%29)
Rate = ![\frac{1}{v^3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bv%5E3%7D)
We were also told that the in the reaction, the gaseous system has an initial volume of 3.00 L and rate of formation of 0.0120 Ms⁻¹
So we can have:
0.0120 = ![\frac{1}{3^3}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%5E3%7D)
0.0120 =
-----Equation (1)
Now; the new rate of formation when the volume of the system decreased to 1.00 L can now be calculated as:
x = ![(\frac{1}{1})^3](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B1%7D%29%5E3)
x = 1 ------- Equation (2)
Dividing equation (2) with equation (1); we have:
= ![\frac{\frac{1}{27} }{1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B1%7D%7B27%7D%20%7D%7B1%7D)
= ![\frac{1}{27}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B27%7D)
x = 0.0120 × 27
x = 0.324 M s⁻¹
∴ the new rate of formation of NOCl = 0.324 M s⁻¹